Round Robin Scheduling Program - c

Question

I have been working on a Round Robin Scheduling Program. My inputs are:
Process Arrival Time Burst Time
1 0 4
2 2 2
3 4 3
4 6 5
5 7 1
Time Slice is 3 units!
My output must be:
Process AT BT WT TT FT
1 0 4 9 13 13
2 2 2 1 3 5
3 4 3 1 4 8
4 6 5 4 9 15
5 7 1 4 5 12
But I am not getting the correct results(WT & FT) for Process 1, 4 & 5. Here's my code, can anyone please help me fix it and get the above results?
#include<stdio.h>
#include<conio.h>
struct proc
{
int id;
int arrival;
int burst;
int rem;
int wait;
int finish;
int ti;
int turnaround;
float ratio;
}process[10];
int no,k;
int chkprocess(int);
void main()
{
int i,j,t,time = 0,n;
struct proc temp;
int nextprocess(int);
clrscr();
printf("\n \n Enter the number of processes: ");
scanf("%d", &n);
printf("\n \n Enter the time slice of the CPU: ");
scanf("%d", &t);
for(i = 1; i <= n; i++)
{
process[i].id = i;
printf("\n\nEnter the arrival time for process %d: ", i);
scanf("%d", &(process[i].arrival));
printf("\nEnter the burst time for process %d: ", i);
scanf("%d", &(process[i].burst));
process[i].rem = process[i].burst;
process[i].ti=0;
process[i].wait=0;
process[i].finish=0;
}
for(i = 1; i <= n; i++)
{
for(j = i + 1; j <= n; j++)
{
if(process[i].arrival > process[j].arrival)
{
temp = process[i];
process[i] = process[j];
process[j] = temp;
}
}
}
no = 0;
j = 1;
while(chkprocess(n) == 1)
{
if(process[no + 1].arrival == time)
no++;
if((process[j].ti<=t)&&(process[j].rem !=0))
{
process[j].rem--;
process[j].ti++;
for(i = 1; i <= no; i++)
{
if((i!=j) && (process[i].rem != 0))
process[i].wait++;
}
}
if(process[j].rem==0)
process[j].finish=time;
if((process[j].ti >= t)||(process[j].rem==0))
{
process[j].ti = 0;
j=nextprocess(j);
}
time++;
}
process[n].finish = time;
printf("\n\n Process Arrival Burst Waiting Finishing turnaround Tr/Tb \n");
printf("%5s %9s %7s %10s %8s %9s\n\n", "id", "time", "time", "time", "time", "time");
for(i = 1; i <= n; i++)
{
process[i].turnaround = process[i].wait + process[i].burst;
process[i].ratio = (float)process[i].turnaround / (float)process[i].burst;
printf("%5d %8d %7d %8d %10d %9d %10.1f ", process[i].id, process[i].arrival,
process[i].burst,
process[i].wait, process[i].finish,
process[i].turnaround, process[i].ratio);
printf("\n\n");
}
getch();
}
int chkprocess(int s)
{
int i;
for(i = 1; i <= s; i++)
{
if(process[i].rem != 0)
return 1;
}
return 0;
}
int nextprocess(int k)
{
int i;
i=k+1;
while(chkprocess(i) && i!=k)
{
if(process[i].rem != 0)
return i;
else
i=(i+1)%no;
}
}
Thank You

Answer 1

I'm sure there are many bugs (starting with not caring if the user wants to enter 11 or more processes even though your array of processes is limited to 10).
However; I spent 10 minutes trying to decipher your code and still don't really know what it thinks it's doing - there's no comments at all and the variable names and function names don't help (e.g. no is not a boolean "yes/no" variable, checkprocess() doesn't check one process but checks all processes to see if all processes have finished, etc). Mostly, if I were being paid to fix this code I'd simple throw it out and rewrite it from scratch to save time. I thought about rewriting it from scratch and just posting the resulting code; but that's not going to help you with your homework.
My advice is, rewrite it from scratch instead of fixing it.
It should have a global currently_running_process variable, a global current_time variable, one function to increase the current time, and one function for the scheduler itself.
The function to increase the current time would:
for each process on the scheduler's linked list, increase the waiting time
do current_time++
find any processes that should be started (current_time == arrival_time) and append any started processes to the end of the scheduler's linked list
The scheduler function should:
remove the first process from the scheduler's linked list
determine how much time that process should use (the time slice length or the process' remaining time, whichever is lower)
subtract that amount of time from the process' remaining time
call the increase_time() function in a loop, until that amount of time has passed
if the process' remaining time is not zero; put the process back onto the end of the linked list
if the process` remaining time was zero, check if the scheduler's linked list is empty and exit the program if it is
Note: I'd start with current_time = -1; and call the function to increase the current time once before calling the scheduler function; so that any processes with arrival_time == 0 would be added to the scheduler's linked list before the scheduler starts working (and so that the scheduler function doesn't see an empty list as soon as it's started).

Answer 2

/* The following code doesn't take the arrival time of the processes in account.
HAPPY CODING */
#include<stdio.h>
void main()
{
int b[10],br[10],wo[10];
int n,i,bt,q,count;
float awt=0,att=0;
for (i=0;i<10;i++)
wo[i]=0;
printf("Input the nmbr of processes running....");
scanf("%d",&n);
printf("\n Input their burst tym in order..");
for(i=0;i<n;i++)
scanf("%d",&b[i]);
printf("\n Input the quantum time for the algorithm..");
scanf("%d",&q);
for(i=0;i<n;i++)
br[i]=b[i];
bt=0;
for(i=0;i<n;i++)
bt=bt+b[i];
count=0;
printf("\nThe Gantt Chart is as follows:\n");
printf("\n 0");
do
{
for(i=0;i<n;i++)
{
if(br[i]==0)
{}
else
{
if(br[i]>=q)
{
br[i]=br[i]-q;
if(br[i]==0)
wo[i]=count;
count=count+q;
printf("\t(P%d)",i);
printf("\t%d",count);
}
else
{
if(br[i]<q)
{
count=count+br[i];
br[i]=0;
wo[i]=count;
printf("\t(P%d)",i);
printf("\t%d",count);
}
}
}
}
}while(count<bt);
for(i=0;i<n;i++)
awt=awt+(wo[i]-b[i]);
awt=awt/n;
printf("\n The average waiting time is....%f",awt);
for(i=0;i<n;i++)
att=att+wo[i];
att=att/n;
printf("\n The average turnaround time is....%f",att);
}

Answer 3

You can use queue for doing the same, i am pasting a link which is written in ANSI CPP
You can check this link for more info. I was having same problem like you had but the code on the link helped me a lot it also contains many other Scheduling program but i extracted only round robin from it.
click here to see the code for round robin Scheduling

Answer 4

Round Robin Scheduling program in C
#include<stdio.h>
#include<conio.h>
main(){
int i, j, k, n, so, tq, sob, sum, swt, stat, tata, temp, count;
int bt[10], bth[10], wt[10], tat[10];
float awt=0.0, atat=0.0;
char new;
// i = loop controller
// j = loop controller
// k = loop controller
// n = number of process
// so = (burst time holder divided by time quantum) and added by one
// tq = time quantum
// awt =average waiting time
// new = hold the value of start command
// sob = gantt chart size from so
// swt = summation of waiting time l
// bt[] = burst time
// wt[] = waiting time
// atat = average turn around time
// gcps = gantt chart process sequence
// stat = summation of turn around time
// tata = accumulator of turn around time
// temp = time quantum holder
// count = counter
// bth[] = burst time holder
// tat[] = turn around time
printf("\n\n\n\n To start round robin scheduling press any key: ");
k = 0;
new = getche();
system("cls");
while(k < 7){
j = 0; sob = 0; count = 0; sum = 0; swt = 0; stat = 0; tata = 0;
printf("\n\n\n\t\t\t ROUND-ROBIN SCHEDULING");
printf("\n\t\t\t ======================");
printf("\n\n\n\n\n Enter number of processes: ");
scanf("%d", &n);
printf("\n");
for(i = 0; i < n; i++){
printf("\n Enter burst time for Process P%d: ", i+1);
scanf("%d", &bt[i]);
bth[i] = bt[i];
}
printf("\n\n Enter time quantum: ");
scanf("%d", &tq);
system("cls");
printf("\n\n\n\t\t\t ROUND-ROBIN SCHEDULING");
printf("\n\t\t\t ======================");
printf("\n\n\n\n\n Time quantum: %d", tq);
for(i = 0; i < n; i++){
if(bth[i] % tq == 0){
so = bth[i] / tq;
}
else{so = (bth[i] / tq) +1;}
sob = sob + so;
}
int gc[sob], gcps[sob];
while(1){
for(i = 0,count = 0; i < n; i++){
temp = tq;
if(bth[i] == 0){
count++;
continue;
}
if(bth[i] > tq){
gc[j] = tq;
gcps[j] = i+1; j++;
bth[i] = bth[i] - tq;
}
else if(bth[i] >= 0){
if(bth[i] == tq){gc[j] = tq; gcps[j] = i+1; j++;}
else{gc[j] = bth[i]; gcps[j] = i+1; j++;}
temp = bth[i];
bth[i] = 0;
}
tata = tata + temp;
tat[i ]= tata;
}
if(n==count){
break;
}
}
for(i = 0; i < n; i++){
wt[i] = tat[i] - bt[i];
swt = swt + wt[i];
stat = stat + tat[i];
}
awt = (float)swt/n;
atat = (float)stat/n;
printf("\n\n Process Burst time Waiting time Turn around time\n");
printf(" ------- ---------- ------------ ----------------\n");
for(i = 0; i < n; i++){
printf("\n\n P%d\t %d\t %d \t %d", i+1, bt[i], wt[i], tat[i]);
}
printf("\n\n\n\n Gantt Chart:\n");
printf(" ------------\n\n");
for(j = 0; j < sob; j++){
printf("\tP%d", gcps[j]);
}
printf("\n 0");
for(j = 0; j < sob; j++){
sum = sum + gc[j];
if(j == 0){printf(" %d", sum);}
else{printf("\t %d", sum);}
}
printf("\n\n\n\n Average waiting time: %.2f \n\n Average turn around time: %.2f",awt,atat);
printf("\n\n\n\n To start again press S and to exit press any key: ");
new = getche();
system("cls");
if(new == 'S'|| new == 's'){k++;}
else{printf("\n\n\n Program was terminated successfully\n\n Thank you\n\n\n"); break;}
}
}

Answer 5

This code will read data from file whose format should have one process info in a single line, arrival time, burst time, spaced, and file should terminate with -1. File name and time slice must be passed in command arguments. Like:
0 3
1 2
2 1
-1
The code is in C and variable names are self-descriptive.
#include<stdio.h>
int main(int argc, char *argv[])
{
int flag = 0;
int timeSlice = atoi(argv[2]);
printf("%d\n\n", timeSlice);
int arrivalTime[10], burstTime[10], responseTime[10], finishTime[10];
int remainingProcesses, processCount = 0;
FILE * file = fopen(argv[1], "r");
if (!(file == NULL))
{
while (fscanf(file, "%d", &arrivalTime[processCount]))
{
if (arrivalTime[processCount] == -1)
break;
fscanf(file, "%d", &burstTime[processCount]);
responseTime[processCount] = burstTime[processCount];
processCount++;
}
remainingProcesses = processCount;
fclose(file);
}
printf("Process\t| Arrival time\t| Finish Time\t| Burst\t| Turnaround\t|\n");
printf("-------------------------------------------------------------------------\n");
int i = 0; int time = 0;
while (remainingProcesses != 0)
{
if (responseTime[i] <= timeSlice && responseTime[i]>0)
{
time += responseTime[i];
responseTime[i] = 0;
flag = 1;
}
else if (responseTime[i] > 0)
{
responseTime[i] -= timeSlice;
time += timeSlice;
}
if (responseTime[i] == 0 && flag == 1)
{
finishTime[i] = time;
remainingProcesses--;
printf("P[%d]\t|\t%d\t|\t%d\t|\t%d\t|\t%d\t|\n", i + 1, arrivalTime[i], finishTime[i], burstTime[i], finishTime[i] - arrivalTime[i]);
flag = 0;
}
if (i == processCount - 1) // If its the last process go back to slicing process 1
{
i = 0;
}
else if (arrivalTime[i + 1] <= time) // If the next process has kicked in
{
i++;
}
else // If the process haven't kicked in yet
{
time++;
i = 0;
}
}
return 0;
}

Answer 6

specials thank to kanika
I just modified some code.
output:
#include<stdio.h>
void main()
{
int bt[10],btTmp[10],tat[10];
int n,i,btAll,quantum,count;
float awt=0,att=0;
for (i=0;i<10;i++)
tat[i]=0;
printf("Input the number of processes : ");
scanf("%d",&n);
printf("\n Input their burst time in order : ");
for(i=0;i<n;i++){
scanf("%d",&bt[i]);
btTmp[i]=bt[i];
}
printf("\n Input the quantum time : ");
scanf("%d",&quantum);
btAll=0;
for(i=0;i<n;i++)
btAll=btAll+bt[i];
count=0;
while(count<btAll)
{
for(i=0;i<n;i++)
{
if(btTmp[i]==0)
{}
else
{
if(btTmp[i]>=quantum)
{
btTmp[i]=btTmp[i]-quantum;
if(btTmp[i]==0)
tat[i]=count+quantum;
count=count+quantum;
}
else
{
count=count+btTmp[i];
btTmp[i]=0;
tat[i]=count;
}
}
}
}
printf("\nprocess \t bt time \t wt time \t TAT\n");
for(i=0;i<n;i++){
printf(" \tP%d \t\t %d \t\t %d \t \t %d\n",i, bt[i], tat[i]-bt[i], tat[i]);
awt=awt+(tat[i]-bt[i]);
att=att+tat[i];
}
printf("\n The average waiting time is %f : ",awt/n);
printf("\n The average turnaround time is....%f : ",att/n);
}